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An organic compound was analysed by duma...

An organic compound was analysed by dumas method. 0.45 gm of the compound on combustion gave 48.6 ml nitrogen at `27^@C` and 756 mm pressure. Calculate the percentage composition of the compound.

Text Solution

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`V_(1)=48.6mL" "V_(2)=?`
`P_(1) = 756 mm" "P_(2) = 760 mm`
`T_(1) = 27 +273" "T_(2) = 273 K = 300K`
Applying general gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
Volume of nitrogen at `STP, V_(2) = (P_(1)V_(1))/(T_(1)).(T_(2))/(P_(2))`
`= (756 xx 48.6)/(300) xx (273)/(760)`
`= 43.99mL`
Mass of organic compound `= 0.45g`
Percentage of nitrogen in the compound `= (28)/(22400) xx (V_(2))/(W) xx 100`
`= (28)/(22400) xx (43.99)/(0.45) xx 100`
`=12.22%`
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